Evaluate the improper integral if it exists. $\int_{1}^{\infty}\dfrac{1}{x^3}\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac14$ (Choice B) B $\dfrac12$ (Choice C) C $1$ (Choice D) D The improper integral diverges.
Explanation: First, let's rewrite the improper integral: $\int_{1}^{\infty}\dfrac{1}{x^3}\,dx =\lim_{b\to\infty}\int_1^{b} \dfrac{1}{x^3}\,dx$ We can now evaluate the integral: $\begin{aligned} \phantom{\int_1^{\infty}\dfrac1{x^3}\,dx}&=\lim_{b\to\infty}\int_1^b \dfrac1{x^3}\,dx\\ \\ \\ &=\lim_{b\to\infty}\left[-\dfrac1{2x^2}\right]_1^b\\ \\ \\ &=\lim_{b\to\infty}\left(\dfrac12-\dfrac1{2b^2}\right)\\ \\ \\ &=\lim_{b\to\infty}\left(\dfrac12\right)-\lim_{b\to\infty}\left(\dfrac1{2b^2}\right)\\\\ &=\dfrac12-0\\ \\ &=\dfrac12 \end{aligned}$ The answer: $\int_{1}^{\infty}\dfrac{1}{x^3}\,dx =\dfrac12$